Convert uint8_t hex value to binary

Question

So like it says in the title I am trying to convert a hexadecimal into a binary. But the problem I have been facing is that, the given value is an uint32_t type. So far, I have convert from uint32_t to uint8_t such each of the uint8_t variables hold 8 bits of the original hex value. I have succesfully converted the hexadecimal into binary, in the code shown below. But now i am unable to print it in a certain format.

    The Format is:
    1101 1110 1010 1101 1011 1110 1110 1111
    ---0 1010 1101 1011 1110 1110 1111
    --01 1011 1110 1110 1111


    void convert(uint32_t value,bool bits[],int length)
    {
          uint8_t a0,a1,a2,a3;
          int i,c,k,j;
          a0 = value;
          a1 = value >> 8;
          a2 = value >> 16;
          a3 = value >> 24;
          for(i=0,c=7;i<=7,c>=0;i++,c--)
          {
                k = a3>>c;
                if(k&1)
                   bits[i] = true;
                else
                   bits[i] = false;
          }
          for(i=8,c=7;i<=15,c>=0;i++,c--)
          {
               k = a2>>c;
               if(k&1)
                  bits[i] = true;
               else
                  bits[i] = false;
           } 
          for(i=16,c=7;i<=23,c>=0;i++,c--)
          {
                k = a1>>c;
               if(k&1)
                   bits[i] = true;
               else
                   bits[i] = false;
          }
          for(i=24,c=7;i<=31,c>=0;i++,c--)
          {
                k = a0>>c;
                if(k&1)
                    bits[i] = true;
                else
                    bits[i] = false;
          }
          for(i=0;i<32;i=i+4)
          {
               for(j=i;j<i+4;j++)
               {
                    printf("%d",bits[j]);
               }
               printf(" ");
           }
     }
     void printBits(bool *bits,int length)
     {
          int len =32,i,j;
          int y = len-length-3;
          printf("\n");
          for(i=0,j=y;i<32,j<32;i++,j++)
          {
            // Trying to come up with some idea
          }
     }
     int main( int argc, const char * argv[] )
     {
            uint32_t value = 0xDEADBEEFL;
            bool     bits[32];
            int len;              
            for (  len = 32; len > 0; len -= 7 )
            {
                 convert (value, bits, len );
                 printBits (bits, len);
            }
            return 0;
     }

Even though I think the main idea is that on every iteration len out of 32 bits must be converted into binary and printed, I was not able to incorporate that idea into my code. So I decided to convert the entire hex number and try to print only what is needed.

Answer 1

Here some solutions!!!

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

void toBinary(uint8_t a)
{
    uint8_t i;

    for(i=0x80;i!=0;i>>=1)
        printf("%c",(a&i)?'1':'0'); 
}

void ul32toBinary(uint32_t a)
{
    uint32_t i;

    for(i=0x80000000;i!=0;i>>=1)
        printf("%c",(a&i)?'1':'0'); 
}

int main(void)
{
    uint32_t k=0xDEADBEEF;

    uint8_t *x;

    int i;

    printf("Converting 0xCC:\t");
    toBinary(0xCC); puts("");   

    printf("Converting 0xF1:\t");
    toBinary(0xF1); puts("");

    printf("Converting 0x%08X:\t",k);
    ul32toBinary(k);puts("");   

    printf("Converting 0x%08X:\t",k);
    x=(char *)&k;   
    //This is valid on little endian CPUs as Intel are!
    for(i=3;i>=0;i--) {
        toBinary(*(x+i));
        printf(" ");            
    }
    puts("");

    return 0;
}

This is the ouput of this code!

Converting 0xCC:    11001100
Converting 0xF1:    11110001
Converting 0xDEADBEEF:  11011110101011011011111011101111
Converting 0xDEADBEEF:  11011110 10101101 10111110 11101111 

To have the ouput you indicate as EXPECTED, you have to slightly modify the function toBinary() in the code above. Remember that a byte is 8 bits, not 4! To have the format you want, just insert an "if" in the right place and write a space ... an hint is: if (i==0x10) {} ... !!!

Goodnight :)

The following code is the code to print a byte in two nibbles of 4 bits:

void toBinary2(uint8_t a)
{
    uint8_t i;

    for(i=0x80;i!=0;i>>=1) {
        printf("%c",(a&i)?'1':'0');
        if (i==0x10)
            printf(" ");
    }
}

Answer 2

You're shifting too far for one thing...i is going 8..1 where it should go 7..0 (and i should be, like all loop counters, an int. If its unsigned like you have it you won't ever be < 0 to stop the loop). Also, you're unpacking the 32-bit value wrong:

a0 = (value >> 24) & 0xFF;
a1 = (value >> 16) & 0xFF;
a2 = (value >> 8) & 0xFF;
a3 = value & 0xFF;